NAME
Math::PlanePath::TerdragonCurve  triangular dragon curve
SYNOPSIS
use Math::PlanePath::TerdragonCurve;
my $path = Math::PlanePath::TerdragonCurve>new;
my ($x, $y) = $path>n_to_xy (123);
DESCRIPTION
This is the terdragon curve by Davis and Knuth,
30 28 7
/ \ / \
/ \ / \
31,34  26,29,32  27 6
\ / \
\ / \
24,33,42  22,25 5
/ \ / \
/ \ / \
40,43,46  20,23,44  12,21 10 4
\ / \ / \ / \
\ / \ / \ / \
18,45  13,16,19  8,11,14  9 3
\ / \ / \
\ / \ / \
17 6,15  4,7 2
\ / \
\ / \
2,5  3 1
\
\
0  1 <Y=0
^ ^ ^ ^ ^ ^ ^ ^
4 3 2 1 X=0 1 2 3
Points are a triangular grid using every second integer X,Y as per "Triangular Lattice" in Math::PlanePath.
The base figure is an "S" shape
23
\
\
01
which then repeats in selfsimilar style, so N=3 to N=6 is a copy rotated +120 degrees, which is the angle of the N=1 to N=2 edge,
6 4 base figure repeats
\ / \ as N=3 to N=6,
\/ \ rotated +120 degrees
5 23
\
\
01
Then N=6 to N=9 is a plain horizontal, which is the angle of N=2 to N=3,
89 base figure repeats
\ as N=6 to N=9,
\ no rotation
67,4
\ / \
\ / \
5,23
\
\
01
Notice X=1,Y=1 is visited twice as N=2 and N=5. Similarly X=2,Y=2 as N=4 and N=7. Each point can repeat up to 3 times. "Inner" points are 3 times and on the edges up to 2 times. The first tripled point is X=1,Y=3 which as shown above is N=8, N=11 and N=14.
The curve never crosses itself. The vertices touch as triangular corners and no edges repeat.
The curve turns are the same as the GosperSide
, but here the turns are by 120 degrees each whereas GosperSide
is 60 degrees each. The extra angle here tightens up the shape.
Spiralling
The first step N=1 is to the right along the X axis and the path then slowly spirals anticlockwise and progressively fatter. The end of each replication is
Nlevel = 3^level
That point is at level*30 degrees around (as reckoned with Y*sqrt(3) for a triangular grid).
Nlevel X, Y Angle (degrees)
  
1 1, 0 0
3 3, 1 30
9 3, 3 60
27 0, 6 90
81 9, 9 120
243 27, 9 150
729 54, 0 180
The following is points N=0 to N=3^6=729 going halfcircle around to 180 degrees. The N=0 origin is marked "0" and the N=729 end is marked "E".
* * * *
* * * * * * * *
* * * * * * * *
* * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * *
* E * * * * * * * * * * * * * * * * 0 *
* * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * *
* * * * * * * *
* * * * * * * *
* * * *
Tiling
The little "S" shapes of the base figure N=0 to N=3 can be thought of as a rhombus
23
. .
. .
01
The "S" shapes of each 3 points make a tiling of the plane with those rhombi
\ \ / / \ \ / /
*** ***
/ / \ \ / / \ \
\ / / \ \ / / \ \ /
** *** **
/ \ \ / / \ \ / / \
\ \ / / \ \ / /
*** ***
/ / \ \ / / \ \
\ / / \ \ / / \ \ /
** *o* **
/ \ \ / / \ \ / / \
\ \ / / \ \ / /
*** ***
/ / \ \ / / \ \
Which is an ancient pattern,
Arms
The curve fills a sixth of the plane and six copies rotated by 60, 120, 180, 240 and 300 degrees mesh together perfectly. The arms
parameter can choose 1 to 6 such curve arms successively advancing.
For example arms => 6
begins as follows. N=0,6,12,18,etc is the first arm (the same shape as the plain curve above), then N=1,7,13,19 the second, N=2,8,14,20 the third, etc.
\ / \ /
\ / \ /
 8/13/31  7/12/30 
/ \ / \
\ / \ / \ /
\ / \ / \ /
 9/14/32  0/1/2/3/4/5  6/17/35 
/ \ / \ / \
/ \ / \ / \
\ / \ /
 10/15/33  11/16/34 
/ \ / \
/ \ / \
With six arms every X,Y point is visited three times, except the origin 0,0 where all six begin. Every edge between points is traversed once.
FUNCTIONS
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
$path = Math::PlanePath::TerdragonCurve>new ()
$path = Math::PlanePath::TerdragonCurve>new (arms => 6)

Create and return a new path object.
The optional
arms
parameter can make 1 to 6 copies of the curve, each arm successively advancing. ($x,$y) = $path>n_to_xy ($n)

Return the X,Y coordinates of point number
$n
on the path. Points begin at 0 and if$n < 0
then the return is an empty list.Fractional positions give an X,Y position along a straight line between the integer positions.
$n = $path>xy_to_n ($x,$y)

Return the point number for coordinates
$x,$y
. If there's nothing at$x,$y
then returnundef
.The curve can visit an
$x,$y
up to three times.xy_to_n()
returns the smallest of the these N values. @n_list = $path>xy_to_n_list ($x,$y)

Return a list of N point numbers for coordinates
$x,$y
. There can be none, one, two or three N's for a given$x,$y
.
Descriptive Methods
$n = $path>n_start()

Return 0, the first N in the path.
$dx = $path>dx_minimum()
$dx = $path>dx_maximum()
$dy = $path>dy_minimum()
$dy = $path>dy_maximum()

The dX,dY values on the first arm take three possible combinations, being 120 degree angles.
dX,dY for arms=1  2, 0 dX minimum = 1, maximum = +2 1, 1 dY minimum = 1, maximum = +1 1,1
For 2 or more arms the second arm is rotated by 60 degrees so giving the following additional combinations, for a total six. This changes the dX minimum.
dX,dY for arms=2 or more  2, 0 dX minimum = 2, maximum = +2 1, 1 dY minimum = 1, maximum = +1 1,1
FORMULAS
N to X,Y
There's no reversals or reflections in the curve so n_to_xy()
can take the digits of N either low to high or high to low and apply what is effectively powers of the N=3 position. The current code goes low to high using i,j,k coordinates as described in "Triangular Calculations" in Math::PlanePath.
si = 1 # position of endpoint N=3^level
sj = 0 # where level=number of digits processed
sk = 0
i = 0 # position of N for digits so far processed
j = 0
k = 0
loop base 3 digits of N low to high
if digit == 0
i,j,k no change
if digit == 1
(i,j,k) = (sij, sjk, sk+i) # rotate +120, add si,sj,sk
if digit == 2
i = sk # add (si,sj,sk) rotated +60
j += si
k += sj
(si,sj,sk) = (si  sk, # add rotated +60
sj + si,
sk + sj)
The digit handling is a combination of rotate and offset,
digit==1 digit 2
rotate and offset offset at si,sj,sk rotated
^ 2>
\
\ \
* 1 * *
The calculation can also be thought of in term of w=1/2+I*sqrt(3)/2, a complex number sixth root of unity. i is the real part, j in the w direction (60 degrees), and k in the w^2 direction (120 degrees). si,sj,sk increase as if multiplied by w+1.
Turn
At each point N the curve always turns 120 degrees either to the left or right, it never goes straight ahead. If N is written in ternary then the lowest nonzero digit gives the turn
ternary lowest
nonzero digit turn
 
1 left
2 right
At N=3^level or N=2*3^level the turn follows the shape at that 1 or 2 point. The first and last unit step in each level are in the same direction, so the next level shape gives the turn.
2*3^k3^(k+1)
\
\
01*3^k
Next Turn
The next turn, ie. the turn at position N+1, can be calculated from the ternary digits of N similarly. The lowest non2 digit gives the turn.
ternary lowest
non2 digit turn
 
0 left
1 right
If N is all 2s then the lowest non2 is taken to be a 0 above the high end. For example N=8 is 22 ternary so considered 022 for lowest non2 digit=0 and turn left after the segment at N=8, ie. at point N=9 turn left.
This rule works for the same reason as the plain turn above. The next turn of N is the plain turn of N+1 and adding +1 turns trailing 2s into trailing 0s and increments the 0 or 1 digit above them to be 1 or 2.
Total Turn
The direction at N, ie. the total cumulative turn, is given by the number of 1 digits when N is written in ternary,
direction = (count 1s in ternary N) * 120 degrees
For example N=12 is ternary 110 which has two 1s so the cumulative turn at that point is 2*120=240 degrees, ie. the segment N=16 to N=17 is at angle 240.
The segments for digit 0 or 2 are in the "current" direction unchanged. The segment for digit 1 is rotated +120 degrees.
X,Y to N
The current code applies TerdragonMidpoint
xy_to_n()
to calculate six candidate N from the six edges around a point. Those N values which convert back to the target X,Y by n_to_xy()
are the results for xy_to_n_list()
.
The six edges are three going towards the point and three going away. The midpoint calculation gives N1 for the towards and N for the away. Is there a good way to tell which edge will be the smaller? Or just which 3 edges lead away? It would be directions 0,2,4 for the even arms and 1,3,5 for the odd ones, but identifying the boundaries of those arms to know which is which is difficult.
X,Y Visited
When arms=6 all "even" points of the plane are visited. As per the triangular representation of X,Y this means
X+Y mod 2 == 0 "even" points
Boundary Length
The length of the boundary of the terdragon on points N=0 to N=3^k inclusive, taking each line segment as length 1, is
boundary[k] = / 2 if k=0 (N=0 to N=1)
\ 3*2^k if k>=1 (N=0 to N=3^k)
= 2, 6, 12, 24, 48, ...
The boundary follows the curve edges around from the origin until returning there. So the single line segment N=0 to N=1 is boundary length 2, or the "S" shape of N=0 to N=3 is length 6.
23
boundary[0] = 2 \
\ boundary[1] = 6
01 01
The boundary[1] first "S" is 3x the length of the preceding but thereafter the way the curve touches itself means the boundary grows by only 2x per level.
The boundary formula can be calculated from the way the curve meets when it replicates. Consider the level N=0 to N=3^k and take its boundary length in two parts as a short side R on the right and the "V" shaped indentation L on the left. These are shown as plain lines here but are wiggly as the curve becomes bigger and fatter.
R R[k] = right side boundary length
23 L[k] = left side boundary length
\ L initial
L \ R[0] = 1
01 L[0] = 2
R boundary[k+1] = 2*R[k] + 2*L[k]
boundary[1] = 6
By symmetry the two sides of the terdragon are the same length, so the total boundary is twice the right side,
boundary[k] = 2*R[k+1]
When the curve is tripled out to the next level N=3^k the boundary length does not triple because the sides marked "===" in the following diagram enclose lengths 2*R and 2*L which would have been boundary, leaving only 4*R and 4*L.
R for k >= 0
*3 R[k+1] = R[k] + L[k] # per 0 to 1
\ L L[k+1] = R[k] + L[k] # per 0 to 2
L \
2=====@
\ / \ R
R \ / \ initial boundary[0] = 2
@=====1 boundary[1] = 6
\ L
L \
0*
R
The two recurrences for R and L are the same, so R[k]=L[k] for k>=1 and hence
R[k+1] = 2*R[k] k >= 1
boundary[k] = 2*boundary[k1] k >= 2
= 3*2^k from initial boundary[1] = 6
Area
The area enclosed by the curve from N=0 to N=3^k inclusive is
area[k] = / 0 if k=0
\ 2*(3^(k1)  2^(k1)) if k >=1
= 0, 0, 2, 10, 38, 130, 422, 1330, 4118, ...
The area can be calculated from the total line segments less the boundary segments. Imagine 1/3 of an equilateral triangle on each side of each line segment
* equilateral triangle
/\ divided into 3 parts
/  \
/  \
/ _*_ \ _*_ 1/3 triangle on
/_ _\ _ _ each side of a
** ** line segment
__ __
*
A line segment which is on the boundary of the curve should count only 1 of its triangles towards the area, not 2. A line segment such as N=0 to N=1 which is isolated is both a left and right boundary segment and so it counts 0 triangles towards the area.
So 3^k line segments which is 2*3^k pairs of triangles, less the boundary ones, and each remaining triangle is 1/3 of an equilateral so
2*3^k  boundary[k]
area[k] =  = 2*(3^(k1) + 2^(k1))
3
This calculation can be made because the inside of the curve always has every edge traversed exactly once and hence always has exactly 3 line segments surrounding each enclosed equilateral triangle.
Area by Replication
The area can also be calculated directly from the replication.
*D
\ A[k] = 2 * A[k1] # AB and CD
\ + 2 * 3^(k2) # centre triangles
Cf  2 * A[k2]/2 # Cf, Be insides
\ / \ + 2 * A[k2]/2 # Ce, Bf outsides
\ / \
eB = 2*A[k1] + 2*3^(k2)
\
\ sum to
A* A[k] = 2*(3^(k1)  2^(k1))
The area enclosed by the end two copies AB and CD are each the area of the preceding level.
The middle two triangles enclose area 2*3^k. But they duplicate the area on the underside of the Cf copy of the curve and the upper side of the Be copy. The terdragon is symmetric on the two sides of the line between its endpoints so the part on the upper side is half the curve, so subtract 2*A[k2]/2.
Then there are 2 similar half curve A[k2]/2 areas on the outer sides of the Bf and Ce segments to be added. The extra and overlapped insides and outsides cancel out.
Area as Rhombus
The area of the curve approaches the area of a rhombus made of two large equilateral triangles between the endpoints.
*N N=3^(k+1)
. \ . side length = sqrt(3)^k
. \ . rhombus area = 2 * side^2 = 2*3^k
O* (area measured in unit triangles)
terdragon 2*(3^k  2^k)
 =  > 1 as k>infinity
rhombus 2*3^k
If the terdragon is reckoned as a fractal with unit length between its endpoints and infinitely smaller wiggles then this ratio is exact, ie. the area of the fractal terdragon is the same as the area of the rhombus.
OEIS
The terdragon is in Sloane's Online Encyclopedia of Integer Sequences as,
http://oeis.org/A080846 (etc)
A080846 next turn 0=left,1=right, by 120 degrees
(n=0 is turn at N=1)
A060236 turn 1=left,2=right, by 120 degrees
(lowest nonzero ternary digit)
A137893 turn 1=left,0=right (morphism)
A189640 turn 0=left,1=right (morphism, extra initial 0)
A189673 turn 1=left,0=right (morphism, extra initial 0)
A038502 strip trailing ternary 0s,
taken mod 3 is turn 1=left,2=right
A189673 and A026179 start with extra initial values arising from their morphism definition. That can be skipped to consider the turns starting with a left turn at N=1.
A026225 N positions of left turns,
being (3*i+1)*3^j so lowest nonzero digit is a 1
A026179 N positions of right turns (except initial 1)
A060032 bignum turns 1=left,2=right to 3^level
A062756 total turn, count ternary 1s
A005823 N positions where total turn == 0, ternary no 1s
A007283 boundary length N=0 to N=3^k for k>=1,
being 3*2^k
A002023 boundary odd levels N=0 to N=3^(2k+1),
or even levels one side N=0 to N=3^(2k),
being 6*4^k
A164346 boundary even levels N=0 to N=3^(2k),
or one side, odd levels, N=0 to N=3^(2k+1),
being 3*4^k
A056182 area enclosed N=0 to N=3^k, being 2*(3^k2^k)
A081956 same
A118004 1/2 area N=0 to N=3^(2k+1), odd levels, 9^n4^n
SEE ALSO
Math::PlanePath, Math::PlanePath::TerdragonRounded, Math::PlanePath::TerdragonMidpoint, Math::PlanePath::GosperSide
Math::PlanePath::DragonCurve, Math::PlanePath::R5DragonCurve
Larry Riddle's Terdragon page, for boundary and area calculations of the terdragon as an infinite fractal http://ecademy.agnesscott.edu/~lriddle/ifs/heighway/terdragon.htm
HOME PAGE
http://user42.tuxfamily.org/mathplanepath/index.html
LICENSE
Copyright 2011, 2012, 2013, 2014 Kevin Ryde
This file is part of MathPlanePath.
MathPlanePath is free software; you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation; either version 3, or (at your option) any later version.
MathPlanePath is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
You should have received a copy of the GNU General Public License along with MathPlanePath. If not, see <http://www.gnu.org/licenses/>.